# 1998 IMO Problems/Problem 2

## Problem

In a competition, there are $a$ contestants and $b$ judges, where $b\ge3$ is an odd integer. Each judge rates each contestant as either “pass” or “fail”. Suppose $k$ is a number such that, for any two judges, their ratings coincide for at most $k$ contestants. Prove that $\frac{k}{a}\ge\frac{b-1}{2b}$.

## Solution

Let $c_i$ stand for the number of judges who pass the $i$th candidate. The number of pairs of judges who agree on the $i$th contestant is then given by

\begin{align*} {c_i \choose 2} + {{b - c_i} \choose 2} &= \frac{1}{2}\left(c_i(c_i - 1) + (b - c_i)(b - c_i - 1) \right) \\ &= \frac{1}{2}\left( c_i^2 + (b - c_i)^2 - b \right) \\ &\geq \frac{1}{2}\left( \frac{b^2}{2} - b \right) \\ &= \frac{b^2 - 2b}{4} \end{align*}

where the inequality follows from AM-QM. Since $b$ is odd, $b^2 - 2b$ is not divisible by $4$ and we can strengthen the inequality to

$${c_i \choose 2} + {{b - c_i} \choose 2} \geq \frac{b^2 - 2b + 1}{4} = \left(\frac{b - 1}{2}\right)^2.$$

Letting $N$ stand for the number of instances where two judges agreed on a candidate, it follows that

$$N = \sum_{i = 1}^a {c_i \choose 2} + {{b - c_i} \choose 2} \geq a \cdot \left( \frac{b - 1}{2} \right)^2.$$

The given condition on $k$ implies that

$$N \leq k \cdot {b \choose 2} = \frac{kb(b - 1)}{2}.$$

Therefore

$$a \cdot \left( \frac{b - 1}{2} \right)^2 \leq \frac{kb(b - 1)}{2},$$

which simplifies to

$$\frac{k}{a} \geq \frac{b - 1}{2b}.$$